Question: What is the extraneous solution to these equations? $\dfrac{x^2 + 3}{x + 9} = \dfrac{15x - 51}{x + 9}$
Solution: Multiply both sides by $x + 9$ $ \dfrac{x^2 + 3}{x + 9} (x + 9) = \dfrac{15x - 51}{x + 9} (x + 9)$ $ x^2 + 3 = 15x - 51$ Subtract $15x - 51$ from both sides: $ x^2 + 3 - (15x - 51) = 15x - 51 - (15x - 51)$ $ x^2 + 3 - 15x + 51 = 0$ $ x^2 + 54 - 15x = 0$ Factor the expression: $ (x - 6)(x - 9) = 0$ Therefore $x = 6$ or $x = 9$ The original expression is defined at $x = 6$ and $x = 9$, so there are no extraneous solutions.